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Chapter 6: Problem 38

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Use a graphing utility to graph the hyperbola and its asymptotes. $$9 x^{2}-y^{2}+54 x+10 y+55=0$$

Short Answer

Expert verified
The center of the hyperbola is \((-3, 5)\), the vertices are approximately \((-2.67, 5)\) and \((-3.33, 5)\), the foci are approximately \((-1.95, 5)\) and \((-4.05, 5)\), and the equations of the asymptotes are \(y = 3x + 14\) and \(y = -3x - 4\).

Step by step solution

01

Tranformation to Standard Form

Firstly, we rewrite the given equation by grouping the \(x\) and \(y\) terms together, and moving the constant to the right side: \(9x^{2} + 54x - y^{2} + 10y = -55\). Then, to complete the square for \(x\), we add \((54/2*9)^{2} = 9^{2} = 81\) to both sides. Similarly, for \(y\) we subtract \((10/2*1)^{2} = 5^{2} = 25\) from both sides. The equation then becomes: \( 9(x^{2} + 6x + 9) - (y^{2} - 10y + 25) = -55 + 81 - 25\). This reduces to \(9(x + 3)^{2} - (y - 5)^{2} = 1\). After dividing by 1, the standard form of the equation is \((x + 3)^{2}/(1/9) - (y - 5)^{2}/1 = 1\). So, \(a^{2} = 1/9, b^{2} = 1\). The standard form of the equation of the hyperbola is \((x-h)^{2}/a^{2} - (y-k)^{2}/b^{2} = 1\).
02

Identify the Center, Vertices and Foci

From the standard form of the equation, we can identify center of the hyperbola as \((-3, 5)\), and the values of \(a\) and \(b\) are \(a = 1/3\) and \(b = 1\). The vertices are \((-3 \pm a, 5)\), so they are \((-3 + 1/3, 5)\) and \((-3 - 1/3, 5)\), or approximately \((-2.67, 5)\) and \((-3.33, 5)\). The foci are \((-3 \pm c, 5)\), where \(c = \sqrt{a^{2} + b^{2}} = \sqrt{(1/3)^{2} + 1^{2}} = \sqrt{1 + 1/9} = \sqrt{10/9}\) = approx 1.05. So the foci are approximately \((-3 + 1.05, 5)\) and \((-3 - 1.05, 5)\) which gives us \((-1.95, 5)\) and \((-4.05, 5)\).
03

Compute the Equations of the Asymptotes

The equations of the asymptotes of the hyperbola are given by \(y = k \pm (b/a)(x - h)\). So the equations of the asymptotes are \(y = 5 \pm (1/(1/3))(x + 3)\), which simplifies to \(y = 5 \pm 3(x + 3)\). Thus, the equations of the asymptotes are \(y = 5 + 3x + 9 = 3x + 14\) and \(y = 5 - 3x - 9 = -3x - 4\).
04

Graphing the Hyperbola

At this point, use a graphing utility to plot the hyperbola and its asymptotes. The hyperbola is centered at \((-3, 5)\), it extends along the x-axis by \(1/3\) for the vertices, and the asymptotes are given by the lines \(y = 3x + 14\) and \(y = -3x - 4\).

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